Question

$\int\limits^1_0 x(1-x)³/² dx=4/35 ,Prove it.$

Answer 1

Answer:

$\bf\:\int\limits^1_0\:x(1-x)^{\frac{3}{2}}\:dx=\frac{4}{35}$

Step-by-step explanation:

$Let\:I=\int\limits^1_0\:x(1-x)^{\frac{3}{2}}\:dx$

using the property

$\boxed{\int\limits^a_0\:f(x)\:dx=\int\limits^a_0\:f(a-x)\:dx}$

$\implies\:I=\int\limits^1_0\:(1-x)(1-(1-x))^{\frac{3}{2}}\:dx$

$\implies\:I=\int\limits^1_0\:(1-x)x^{\frac{3}{2}}\:dx$

$\implies\:I=\int\limits^1_0\:(x^{\frac{3}{2}}-x^{\frac{5}{2}})\:dx$

$\implies\:I=[\frac{x^{\frac{5}{2}}}{\frac{5}{2}}}-\frac{x^{\frac{7}{2}}}{\frac{7}{2}}}]\limits^1_0$

$\implies\:I=2[\frac{x^{\frac{5}{2}}}{5}-\frac{x^{\frac{7}{2}}}{7}]^1_0$

$\implies\:I=2[\frac{1^{\frac{5}{2}}}{5}-\frac{1^{\frac{7}{2}}}{7}]-2[0]$

$\implies\:I=2[\frac{1}{5}-\frac{1}{7}]$

$\implies\:I=2[\frac{7-5}{35}]$

$\implies\:I=2[\frac{2}{35}]$

$\therefore\boxed{\bf\:I=\frac{4}{35}}$